HDU 1867 A + B for you again (KMP)

#include <stdio.h>
#include <string.h>

int next[100005];

void build_next(char str[])
{
int i=0,j=-1;
next[0] = -1;
int len2=strlen(str);
while (i < len2)
{
if (j==-1 || str[i] == str[j])
{
i++;
j++;
if (str[i] != str[j])
{
next[i] = j;
}
else
next[i] = next[j];
}
else
j = next[j];
}
}

int KMP(char str1[],char str2[])
{
int  len1=strlen(str1);
int len2=strlen(str2);
build_next(str2);
int i=0,j=0,cnt=0;

while (i < len1 &&j <len2)
{
if (j==-1 || str1[i] == str2[j])
{
i++;
j++;
}
else
j = next[j];
}
if(i==len1)
return j;
return 0;
}

int main()
{
int N,n,i,m;
char str1[100005],str2[100005];
while (~scanf("%s%s",str1,str2))
{
//		gets(str2);                        //gets在这里不行，有空格
n=KMP(str1,str2);
m=KMP(str2,str1);	      //还要考虑模式串的问题，两个字符串都可以作为模式串，所以要有两个KMP函数
if(m == n)
{
if(strcmp(str1,str2)>0)
{
printf("%s",str2);
printf("%s\n",str1+n);
}
else
{
printf("%s",str1);
printf("%s\n",str2+n);
}
}
else  if(n>m)
{
printf("%s",str1);
printf("%s\n",str2+n);
}
else
{
printf("%s",str2);
printf("%s\n",str1+m);
}
}
return 0;
}

。。。。。。。。。。。。。。。。。。。。。。。。。。。。

分析：主要是看清题目的意思。

asdf 和 sdfg asdf的后缀和sdfg的前缀有相同的sdf 所以在相加之后等于： asdfg

asdf ghjk 之间没有共同的前后缀，所以加起来等于 asdfghjk

还要考虑模式串的问题，两个字符串都可以作为模式串，所以要有两个KMP函数

题目最后说guarantee the shortest string first, then the minimum lexicographic second, the same rules for other additions.

输出的时候还要满足前面的字符串的字典序要小于后面的字典序

比如 asdfa 和 asdf 首先 n=m。其次因为 strcmp(str1,str2) >0 所以要先输出str2 再输出 str1+n；

A + B for you again

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4884 Accepted Submission(s): 1237



Problem Description

Generally speaking, there are a lot of problems about strings processing. Now you encounter another such problem. If you get two strings, such as “asdf” and “sdfg”, the result of the addition between them is “asdfg”, for “sdf” is the tail substring of “asdf”
and the head substring of the “sdfg” . However, the result comes as “asdfghjk”, when you have to add “asdf” and “ghjk” and guarantee the shortest string first, then the minimum lexicographic second, the same rules for other additions.

Input

For each case, there are two strings (the chars selected just form ‘a’ to ‘z’) for you, and each length of theirs won’t exceed 10^5 and won’t be empty.

Output

Print the ultimate string by the book.

Sample Input

asdf sdfg
asdf ghjk

Sample Output

asdfg
asdfghjk

Author

Wang Ye

Source

2008杭电集训队选拔赛——热身赛